How to calculate Pythagorean intervals

The divisions of the 1st order split the octave into twelve fifths spread over seven octaves with a leftover of 531441/524288 called the Pythagorean comma, approximately equal to a fifty-third of an octave.

The divisions of the 2nd order split the octave into fifty-three intervals spread over [how many?] octaves with a leftover of [ratio], approximately equal to a [n-th] of an octave.

The divisions of the 3rd order split the octave into [how many?] intervals spread over [how many?] octaves with a leftover of [ratio], approximately equal to a [m-th] of an octave.

How can the missing numbers be calculated? Is there a spreadsheet technique or a script or maybe a way to use Scala that could provide not only the missing numbers from above, but also the numbers up to any custom number of orders?

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote

>The divisions of the 1st order split the octave into twelve fifths spread over seven octaves with a leftover of > 531441/524288 called the Pythagorean comma, approximately equal to a fifty-third of an octave.

> The divisions of the 2nd order split the octave into fifty-three intervals spread over [how many?=200]
> octaves with a leftover of [ratio=253], approximately equal to a [n-th] of an octave.

> The divisions of the 3rd order split the octave into [how many?=306] intervals spread over [how >
> many?=665] octaves with a leftover of [ratio8286 ], approximately equal to a [m-th=8591] of an octave.

> How can the missing numbers be calculated? Is there a spreadsheet technique or a script or maybe a way > to use Scala that could provide not only the missing numbers from above, but also the numbers up to any > custom number of orders?

for next orders see especially

http://oeis.org/A060528

or at least the first of them more en detail:

http://xenharmonic.wikispaces.com/Comma
3-limit commas 256 / 243 http://xenharmonic.wikispaces.com/256_243 = |8 -5>: limma or Pythagorean minor second (90.225 cent http://xenharmonic.wikispaces.com/cents)
2187 / 2048 http://xenharmonic.wikispaces.com/2187_2048 = |-11 7>: apotome (113.685 cents)
531441_524288|531441 / 524288 = |-19 12>: Pythagorean comma http://xenharmonic.wikispaces.com/Pythagorean+comma (23.460 cents)
134217728 / 129140163 = |27 -17>: 17-comma http://xenharmonic.wikispaces.com/17-comma (66.765 cents)
70368744177664 / 68630377364883 = |46 -29>: 29-comma http://xenharmonic.wikispaces.com/29-comma (43.305 cents)
36893488147419103232 / 36472996377170786403 = |65 -41>: 41-comma http://xenharmonic.wikispaces.com/41-comma (19.845 cents)
19383245667680019896796723 / 19342813113834066795298816 = |-84 53>: Mercator's comma http://xenharmonic.wikispaces.com/Mercator%27s+comma, 53-comma (3.615 cents)
|485 -306>: Sasktel comma http://xenharmonic.wikispaces.com/Sasktel+comma (1.770 cents)
|-1054 665>: Satanic comma http://xenharmonic.wikispaces.com/Satanic+comma (0.076 cents)

......

http://launch.dir.groups.yahoo.com/neo/groups/tuning/conversations/topics/64423

hope that links do help to answer the initial question
bye
Andy

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

> How can the missing numbers be calculated?

ups,
i forgot to mention:
http://oeis.org/A060528
http://oeis.org/A005664
http://www.eddaardvark.co.uk/t3a1/int_log32.htm

sorry
Andy

baros_ilogic wrote:

> The divisions of the 1st order split the octave into twelve fifths spread over seven octaves with a leftover of 531441/524288 called the Pythagorean comma, approximately equal to a fifty-third of an octave.

> The divisions of the 2nd order split the octave into fifty-three intervals spread over [how many?] octaves with a leftover of [ratio], approximately equal to a [n-th] of an octave.

> The divisions of the 3rd order split the octave into [how many?] intervals spread over [how many?] octaves with a leftover of [ratio], approximately equal to a [m-th] of an octave.

> How can the missing numbers be calculated? Is there a spreadsheet technique or a script or maybe a way to use Scala that could provide not only the missing numbers from above, but also the numbers up to any custom number of orders?

You can calculate the convergents for the perfect fifth in octave, i.e. for log_2(3/2), which are:
0/1, 1/1, 1/2, 3/5, 7/12, 24/41, 31/53, 179/306, 389/665, 9126/15601, 18641/31867, 46408/79335,
65049/111202, 111457/190537, ...

I think* all pairs of subsequent ratios are farey pairs, i.e. pairs (a/c, b/d) for which |bc - ad| = 1.
If we divide this equation by d (and assume d > 0), we get |c * b/d - a| = 1/d.
This tells us that 12 * 24\41 - 7 = +-1\41 ("the difference between 12 41edo-fifths and 7 octaves
is a 41-edo step up or down"), 41 * 31\53 - 24 = +-1\53, 53 * 179/306 - 31 = +-1\306, and so on...

Instead of using only convergents, you can also add semiconvergents (all, or only those that
give better approximations than previous ones), or you can remove some convergents if the
remaining convergents still make an uninterrupted chain of farey pairs. E.g. you can remove
24/41, because 7/12 and 31/53 are still a farey pair, with 12 * 31\53 - 7 = +-1.

Best
Geddy

* can somebody confirm this?

Thanks Andy and Geddy. I'm looking for a symmetry here, and that could be MOS of Pythagorean as far as I know.

Pythagorean Symmetry is achieved when, after a specific number of steps, the intervals seem to divide the octave evenly. "Evenly" is the keyword here, or "seemingly even" to be more precise.

So far, only 12 and 53 meet this requirement - i.e. the distribution of intervals appear to be even and the sonic distance between any 2 consecutive intervals is "relatively the same".

But the division of the octave (using a single interval) in just one direction is just half of the story. If proceeding in the opposite direction (with 2/3 or 4/3 when justified) we come to the second requirement of Pythagorean Symmetry:

The Comma should be the difference between X fifths/fourths and Y octaves, AND between any 2 pseudo-consecutive intervals generated by a member of the ascending series and a member of the descending series, when justified inside the same octave.

An artificial rule here would be that by tempering out the comma, the tempered tones should fall between the intervals created by the ascending and descending series.

So far, only 12 and 53 meet both requirements, and to figure this out I had to draw a logarithmic number line and represent the intervals.

Coming back to my initial question: is there a way to do these calculations, without having to draw hundreds on intervals on a logarithmic number line?

---In tuning-math@yahoogroups.com, <gedankenwelt94@...> wrote:

baros_ilogic wrote:

> The divisions of the 1st order split the octave into twelve fifths spread over seven octaves with a leftover of 531441/524288 called the Pythagorean comma, approximately equal to a fifty-third of an octave.

> The divisions of the 2nd order split the octave into fifty-three intervals spread over [how many?] octaves with a leftover of [ratio], approximately equal to a [n-th] of an octave.

> The divisions of the 3rd order split the octave into [how many?] intervals spread over [how many?] octaves with a leftover of [ratio], approximately equal to a [m-th] of an octave.

> How can the missing numbers be calculated? Is there a spreadsheet technique or a script or maybe a way to use Scala that could provide not only the missing numbers from above, but also the numbers up to any custom number of orders?

You can calculate the convergents for the perfect fifth in octave, i.e. for log_2(3/2), which are:
0/1, 1/1, 1/2, 3/5, 7/12, 24/41, 31/53, 179/306, 389/665, 9126/15601, 18641/31867, 46408/79335,
65049/111202, 111457/190537, ...

I think* all pairs of subsequent ratios are farey pairs, i.e. pairs (a/c, b/d) for which |bc - ad| = 1.
If we divide this equation by d (and assume d > 0), we get |c * b/d - a| = 1/d.
This tells us that 12 * 24\41 - 7 = +-1\41 ("the difference between 12 41edo-fifths and 7 octaves
is a 41-edo step up or down"), 41 * 31\53 - 24 = +-1\53, 53 * 179/306 - 31 = +-1\306, and so on...

Instead of using only convergents, you can also add semiconvergents (all, or only those that
give better approximations than previous ones), or you can remove some convergents if the
remaining convergents still make an uninterrupted chain of farey pairs. E.g. you can remove
24/41, because 7/12 and 31/53 are still a farey pair, with 12 * 31\53 - 7 = +-1.

Best
Geddy

* can somebody confirm this?

Yahoo won't let me upload photos so here is what I mean:
http://i.imgur.com/dkySimH.png

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

Thanks Andy and Geddy. I'm looking for a symmetry here, and that could be MOS of Pythagorean as far as I know.

Pythagorean Symmetry is achieved when, after a specific number of steps, the intervals seem to divide the octave evenly. "Evenly" is the keyword here, or "seemingly even" to be more precise.

So far, only 12 and 53 meet this requirement - i.e. the distribution of intervals appear to be even and the sonic distance between any 2 consecutive intervals is "relatively the same".

But the division of the octave (using a single interval) in just one direction is just half of the story. If proceeding in the opposite direction (with 2/3 or 4/3 when justified) we come to the second requirement of Pythagorean Symmetry:

The Comma should be the difference between X fifths/fourths and Y octaves, AND between any 2 pseudo-consecutive intervals generated by a member of the ascending series and a member of the descending series, when justified inside the same octave.

An artificial rule here would be that by tempering out the comma, the tempered tones should fall between the intervals created by the ascending and descending series.

So far, only 12 and 53 meet both requirements, and to figure this out I had to draw a logarithmic number line and represent the intervals.

Coming back to my initial question: is there a way to do these calculations, without having to draw hundreds on intervals on a logarithmic number line?

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

> So far, only 12 and 53 meet this requirement
> - i.e. the distribution of intervals appear to be even and the sonic distance between any 2 consecutive
> intervals is "relatively the same".

here the next Pythagorean 3-limit convergent is 665,
that satisfies yours "relatively the same" condition much better than 53:

Just compare the both occuring steps:

1200Cents*ln(2^485/3^306)/ln2 = ~1.769735.....Cents
1200Cents*ln(3^359/2^569)/ln2 = ~18453106....Cents

they only differ from each other among them only by the very tiny:

http://xenharmonic.wikispaces.com/Satanic+comma
"....
Satanic comma http://xenharmonic.wikispaces.com/Satanic+comma
The satanic comma is the difference between 666 perfect fifths (octave-reduced) and a single perfect fifth. Equivalently, it's the difference between 665 perfect fifths (octave-reduced) and the unison -- but that wouldn't be as devilishly intriguing.

This difference is inaudible, at only 0.076 cents. The monzo is |-1054 665>.
"
hence therefore you can also use
http://xenharmonic.wikispaces.com/665edo
"
The 665 equal temperament divides the octave into 665 equal parts of 1.80451 cents each. It is best known for its extremely accurate fifth, only 0.0001 cents flat. 665edo is the denominator of a convergent to log2(3), after 41edo http://xenharmonic.wikispaces.com/41edo, 53edo http://xenharmonic.wikispaces.com/53edo and 306edo http://xenharmonic.wikispaces.com/306edo, and before 15601edo http://xenharmonic.wikispaces.com/15601edo....."

check for that the step #6 in:
http://mathforum.org/kb/plaintext.jspa?messageID=5316965
that corresponds to #9 in:
http://www.ericr.nl/wondrous/cycles.html

See also some older discussions about such high-order-3-limit topic under:
http://groups.yahoo.com/neo/groups/makemicromusic/conversations/topics/11815
http://launch.dir.groups.yahoo.com/neo/groups/tuning/conversations/topics/64423

Hope that helps
bye
Andy

or even in that group here:
http://groups.yahoo.com/neo/groups/tuning-math/conversations/messages/4822
"....
0 + 1/(1 + 1/(1 + 1/(2 + 1/(2 + 1/(3 + 1/(1 + 1/(5 + 1/(2)))))))) = 389/665
...."

bye
Andy

It doesn't matter which system satisfies the "relatively the same" condition better (i.e. 53 better than 12), as long as they satisfy it.

I'm not interested in tempering out commas or which (pythagorean) equal temperament has the most advantages. Using cents has nothing to do with this, ratios or exponents should be enough; also if it doesn't help in finding a pattern, the use of logarithms complicates everything.

There is a symmetrical pattern, when bringing into the same octave the generations of ascending and descending intervals (fifths and fourths) and so far just 12 and 53 satisfy all the requirements. I am aware that there are other generations more practical/useful/consonant but this is not what I'm looking for.

Is there a way of finding if the next "relatively the same" generations occurs after 359 or 665 fifths?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

or even in that group here:
http://groups.yahoo.com/neo/groups/tuning-math/conversations/messages/4822
"....
0 + 1/(1 + 1/(1 + 1/(2 + 1/(2 + 1/(3 + 1/(1 + 1/(5 + 1/(2)))))))) = 389/665
...."

bye
Andy

What you're looking for is probably maximal evenness:

http://xenharmonic.wikispaces.com/Maximal+evenness

If you approximate an edo within a larger one, the result is a maximally even scale,
which is either the same edo (e.g. if you "approximate" 12edo in 24edo), or a MOS
scale where the chroma is a single step, i.e. where the small and the large step
differ by a single step.

An example is approximating 12edo in 53edo, where you get a 12-note scale with
step sizes 4 and 5 (differing by 1). Other examples with a fifth generator are
approximations of 12edo in 17-, 29-, 41-, or 65-edo, or 19-, 31-, 43-edo etc. if you
want to go in a meantone direction.
For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
or also 65, 118 etc.

This is due to the farey pair relation I mentioned earlier. If you're looking for tunings
with a 12-note maximally even MOS generated by stacking fifths, you can take a look
the Stern-Brocot tree, and go to the 7/12 node. If you then advance a step up, and
as many steps as you want down (or first down and then repeatedly up), all ratios
along build a farey pair with 7/12, if I don't err.
This means that you can take the denominator of one of those ratios as your edo,
and build a 12-note scale by stacking fifths, and the result is a maximally even
12-note scale.

A pdf of the Stern-Brocot tree / scale tree can be found on anaphoria:
http://anaphoria.com/sctree.PDF
(7/12 is on page 11)

I can also upload a Stern-Brocot tree implemented as a "mind map" in FreeMind,
which is a little larger, and benefits from the search function.

If you're just interested in maximal evenness, and pythagorean approximations don't
matter, you don't need approximations of log_2(3/2) via semiconvergents or convergents,
and you can even go for scales with another generator or period.

I hope I could help ;)
- Geddy

Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

What you're looking for is probably maximal evenness:

http://xenharmonic.wikispaces.com/Maximal+evenness

If you approximate an edo within a larger one, the result is a maximally even scale,
which is either the same edo (e.g. if you "approximate" 12edo in 24edo), or a MOS
scale where the chroma is a single step, i.e. where the small and the large step
differ by a single step.

An example is approximating 12edo in 53edo, where you get a 12-note scale with
step sizes 4 and 5 (differing by 1). Other examples with a fifth generator are
approximations of 12edo in 17-, 29-, 41-, or 65-edo, or 19-, 31-, 43-edo etc. if you
want to go in a meantone direction.
For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
or also 65, 118 etc.

This is due to the farey pair relation I mentioned earlier. If you're looking for tunings
with a 12-note maximally even MOS generated by stacking fifths, you can take a look
the Stern-Brocot tree, and go to the 7/12 node. If you then advance a step up, and
as many steps as you want down (or first down and then repeatedly up), all ratios
along build a farey pair with 7/12, if I don't err.
This means that you can take the denominator of one of those ratios as your edo,
and build a 12-note scale by stacking fifths, and the result is a maximally even
12-note scale.

A pdf of the Stern-Brocot tree / scale tree can be found on anaphoria:
http://anaphoria.com/sctree.PDF
(7/12 is on page 11)

I can also upload a Stern-Brocot tree implemented as a "mind map" in FreeMind,
which is a little larger, and benefits from the search function.

If you're just interested in maximal evenness, and pythagorean approximations don't
matter, you don't need approximations of log_2(3/2) via semiconvergents or convergents,
and you can even go for scales with another generator or period.

I hope I could help ;)
- Geddy

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier: >
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

> Do you draw a horogram?

Already Isaac Newton did that for the 53 case:
http://www.mtosmt.org/issues/mto.93.0.3/mto.93.0.3.lindley7.gif

attend there the traditional nomenclauture:
http://en.wikipedia.org/wiki/Guidonian_hand#Hexachord_in_Middle_Ages
http://www.medieval.org/emfaq/harmony/hex1.html

For an more modern introduction to that topic see:
http://www.elvenminstrel.com/music/tuning/horagrams/horagram_intro.htm

bye
Andy

Hi Bogdan,

sorry for the late reply!

> Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

> How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

I'm not sure "Pythagorean MOS" is an official term; what I meant are simply the
moments of symmetry that result when stacking perfect fifths. Just like I'd say
"meantone MOS" to describe an MOS that results from stacking meantone fifths.

Sorry, I'm aware that I didn't explain many of the terms I used, I just wanted to
make sure that what I suggested is really the solution you are looking for (or at
least something close) before starting with extensive explanations. ;)

Before I explain what I did I should probably explain some basics about MOS's.
I wanted to write a MOS tutorial anyway, so I guess now would be a good time
to start. I can't say how long it'll take, though...

Hi baros_ilogic,
As you may guess from the replies you've received, a lot is known about this process, and there is more than one number series that is related to this matter. Here's an interesting one:
2,3,5,7,12,17,29,41,53,94,147,200,253,306,359,665 ...
I see that Geddy is planning to say more about MOS; meanwhile here is my answer - I hope it will be helpful.

First, though, I am curious as to exactly how you found 53 as the next number after 12; you say that the leftover (Pythagorean comma) is about 1/53 of an octave, but actually it is nearer 1/51 of an octave, so I don't think you can have used it directly. I guess that you simply prolonged the chain of 5ths until you found a leftover that is "a lot smaller than" the small interval; I think that would work, but you would have a long way to go to get to 665, which I think is the next number. All the numbers in the list I gave above are MOS for the 3/2 interacting with the 2/1, but not all of them have a "small" leftover.

Here's an algorithm: given a MOS with N of interval A and M of interval B, find the next MOS by:
if A is the larger interval, then it's N of interval (A-B) and (N+M) of interval B;
if B is the larger interval, then it's M of interval (B-A) and (N+M) of interval A.
Hence define a new N, A, M and B. Repeat ad infinitum.
Note that (A-B) means the difference in cents or equivalently the quotient A/B if you're using fractions.
Start with the two-note MOS; i.e. 1 of interval 3/2 and 1 of interval 4/3.
You will want to do a further check of "evenness"; I suggest A/B (NB here I do mean a quotient of the cents values) between 3/4 and 4/3.

Steve M.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

Hi Steve,

The way I did it was visually; the process is described in this image: http://i.imgur.com/dkySimH.png
That was my initial attempt; I later found out that 12 is not the division of the 1st order, 2 would be that:

Order # Generating steps

1 2 2 5 3 12 4 53 5 306 6 665 7 15.601 8 31.867 9 190.537 10 10.781.274 11 53.715.833
So my question here was how to determine the next steps, without having to draw 306 tiny logarithmic ticks on a number line. The algorithm should not include cents as a measuring tool; this is strictly ratios.

Your algorithm sound interesting, but I have no way of putting it into practice. I have no programming / technical skills, and I am just looking for something very simple, just like the picture I drew. Maybe something that could be calculated in a spreadsheet?

Bogdan
(baros_ilogic)

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi baros_ilogic,
As you may guess from the replies you've received, a lot is known about this process, and there is more than one number series that is related to this matter. Here's an interesting one:
2,3,5,7,12,17,29,41,53,94,147,200,253,306,359,665 ...
I see that Geddy is planning to say more about MOS; meanwhile here is my answer - I hope it will be helpful.

First, though, I am curious as to exactly how you found 53 as the next number after 12; you say that the leftover (Pythagorean comma) is about 1/53 of an octave, but actually it is nearer 1/51 of an octave, so I don't think you can have used it directly. I guess that you simply prolonged the chain of 5ths until you found a leftover that is "a lot smaller than" the small interval; I think that would work, but you would have a long way to go to get to 665, which I think is the next number. All the numbers in the list I gave above are MOS for the 3/2 interacting with the 2/1, but not all of them have a "small" leftover.

Here's an algorithm: given a MOS with N of interval A and M of interval B, find the next MOS by:
if A is the larger interval, then it's N of interval (A-B) and (N+M) of interval B;
if B is the larger interval, then it's M of interval (B-A) and (N+M) of interval A.
Hence define a new N, A, M and B. Repeat ad infinitum.
Note that (A-B) means the difference in cents or equivalently the quotient A/B if you're using fractions.
Start with the two-note MOS; i.e. 1 of interval 3/2 and 1 of interval 4/3.
You will want to do a further check of "evenness"; I suggest A/B (NB here I do mean a quotient of the cents values) between 3/4 and 4/3.

Steve M.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

Hi Bogdan,
Ok, it can be done in a spreadsheet, I'll post it here, or if that's not possible I can send it to you.
Two questions: why do you reject cents as a tool for calculation; you must at least be using logarithms in your diagrams? And did you really find those other orders above 665 using the diagrammatic method?

Steve.

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

Hi Steve,

The way I did it was visually; the process is described in this image: http://i.imgur.com/dkySimH.png
That was my initial attempt; I later found out that 12 is not the division of the 1st order, 2 would be that:

Order # Generating steps

1 2 2 5 3 12 4 53 5 306 6 665 7 15.601 8 31.867 9 190.537 10 10.781.274 11 53.715.833
So my question here was how to determine the next steps, without having to draw 306 tiny logarithmic ticks on a number line. The algorithm should not include cents as a measuring tool; this is strictly ratios.

Your algorithm sound interesting, but I have no way of putting it into practice. I have no programming / technical skills, and I am just looking for something very simple, just like the picture I drew. Maybe something that could be calculated in a spreadsheet?

Bogdan
(baros_ilogic)

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi baros_ilogic,
As you may guess from the replies you've received, a lot is known about this process, and there is more than one number series that is related to this matter. Here's an interesting one:
2,3,5,7,12,17,29,41,53,94,147,200,253,306,359,665 ...
I see that Geddy is planning to say more about MOS; meanwhile here is my answer - I hope it will be helpful.

First, though, I am curious as to exactly how you found 53 as the next number after 12; you say that the leftover (Pythagorean comma) is about 1/53 of an octave, but actually it is nearer 1/51 of an octave, so I don't think you can have used it directly. I guess that you simply prolonged the chain of 5ths until you found a leftover that is "a lot smaller than" the small interval; I think that would work, but you would have a long way to go to get to 665, which I think is the next number. All the numbers in the list I gave above are MOS for the 3/2 interacting with the 2/1, but not all of them have a "small" leftover.

Here's an algorithm: given a MOS with N of interval A and M of interval B, find the next MOS by:
if A is the larger interval, then it's N of interval (A-B) and (N+M) of interval B;
if B is the larger interval, then it's M of interval (B-A) and (N+M) of interval A.
Hence define a new N, A, M and B. Repeat ad infinitum.
Note that (A-B) means the difference in cents or equivalently the quotient A/B if you're using fractions.
Start with the two-note MOS; i.e. 1 of interval 3/2 and 1 of interval 4/3.
You will want to do a further check of "evenness"; I suggest A/B (NB here I do mean a quotient of the cents values) between 3/4 and 4/3.

Steve M.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

Hi Steve,

Thanks for your interest in the topic. I don't know what to answer about cents... I'm trying to make an idea about the numbers, using simple and intuitive terms. For example, 3^53 * 2^-84 tells me much more than 3,615045866 cents. The place of the minus sign tells me that the comma appears on the left side of the graphic (meaning the comma is the first interval after 1/1) and is the difference between 53 twelfths and 84 octaves, or between 53 fifths and 31 octaves (because 84-53=31). And if there's small numbers we're after, the resulting decimal number 1,002090314 is as useful as its value in cents when spotting the next smaller comma. The next one after Mercator will be 3^-306 * 2^485; the minus sign is now with the 3, meaning that the comma is at the right side of the graphic (the comma is the last interval before 2/1).

Guess my understanding process has a preference for this kind of thinking. For example I couldn't understand MoS and Horograms until I got rid of cents and did all the calculations with ratios, like I explained here: http://groups.yahoo.com/neo/groups/tuning-math/conversations/messages/21381

The diagrams do use logarithms - they are actually ticks on a logarithmic number line. I only found the intervals up to 53 using the diagram method.

Bogdan

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi Bogdan,
Ok, it can be done in a spreadsheet, I'll post it here, or if that's not possible I can send it to you.
Two questions: why do you reject cents as a tool for calculation; you must at least be using logarithms in your diagrams? And did you really find those other orders above 665 using the diagrammatic method?

Steve.

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

Hi Steve,

The way I did it was visually; the process is described in this image: http://i.imgur.com/dkySimH.png
That was my initial attempt; I later found out that 12 is not the division of the 1st order, 2 would be that:

Order # Generating steps

1 2 2 5 3 12 4 53 5 306 6 665 7 15.601 8 31.867 9 190.537 10 10.781.274 11 53.715.833
So my question here was how to determine the next steps, without having to draw 306 tiny logarithmic ticks on a number line. The algorithm should not include cents as a measuring tool; this is strictly ratios.

Your algorithm sound interesting, but I have no way of putting it into practice. I have no programming / technical skills, and I am just looking for something very simple, just like the picture I drew. Maybe something that could be calculated in a spreadsheet?

Bogdan
(baros_ilogic)

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi baros_ilogic,
As you may guess from the replies you've received, a lot is known about this process, and there is more than one number series that is related to this matter. Here's an interesting one:
2,3,5,7,12,17,29,41,53,94,147,200,253,306,359,665 ...
I see that Geddy is planning to say more about MOS; meanwhile here is my answer - I hope it will be helpful.

First, though, I am curious as to exactly how you found 53 as the next number after 12; you say that the leftover (Pythagorean comma) is about 1/53 of an octave, but actually it is nearer 1/51 of an octave, so I don't think you can have used it directly. I guess that you simply prolonged the chain of 5ths until you found a leftover that is "a lot smaller than" the small interval; I think that would work, but you would have a long way to go to get to 665, which I think is the next number. All the numbers in the list I gave above are MOS for the 3/2 interacting with the 2/1, but not all of them have a "small" leftover.

Here's an algorithm: given a MOS with N of interval A and M of interval B, find the next MOS by:
if A is the larger interval, then it's N of interval (A-B) and (N+M) of interval B;
if B is the larger interval, then it's M of interval (B-A) and (N+M) of interval A.
Hence define a new N, A, M and B. Repeat ad infinitum.
Note that (A-B) means the difference in cents or equivalently the quotient A/B if you're using fractions.
Start with the two-note MOS; i.e. 1 of interval 3/2 and 1 of interval 4/3.
You will want to do a further check of "evenness"; I suggest A/B (NB here I do mean a quotient of the cents values) between 3/4 and 4/3.

Steve M.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

Hi Bogdan,
OK I understand; I agree that it's nice to see the powers of 2 and 3, but I can't easily perform the algorithm without cents (see below).
I have put a file called bogdan.xlsx in the Files section of this group; I hope you can read it. The "inputs" are at cells A3:D3, intermediate results below that in columns A:D, and the final results at cells R3:R50. Let me know if you want any further explanation.
Note that column N says if the First Interval is greater than the Second Interval; it is based on the cents values in Columns O and P.
Columns Q and S are alternative attempts to do the same test without using cents, but both fail after row 19. There may be a way to re-arrange the tests so that they work, but I have not spent the time to do it.

Steve.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Hi Steve,

Thanks for your interest in the topic. I don't know what to answer about cents... I'm trying to make an idea about the numbers, using simple and intuitive terms. For example, 3^53 * 2^-84 tells me much more than 3,615045866 cents. The place of the minus sign tells me that the comma appears on the left side of the graphic (meaning the comma is the first interval after 1/1) and is the difference between 53 twelfths and 84 octaves, or between 53 fifths and 31 octaves (because 84-53=31). And if there's small numbers we're after, the resulting decimal number 1,002090314 is as useful as its value in cents when spotting the next smaller comma. The next one after Mercator will be 3^-306 * 2^485; the minus sign is now with the 3, meaning that the comma is at the right side of the graphic (the comma is the last interval before 2/1).

Guess my understanding process has a preference for this kind of thinking. For example I couldn't understand MoS and Horograms until I got rid of cents and did all the calculations with ratios, like I explained here: http://groups.yahoo.com/neo/groups/tuning-math/conversations/messages/21381

The diagrams do use logarithms - they are actually ticks on a logarithmic number line. I only found the intervals up to 53 using the diagram method.

Bogdan

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi Bogdan,
Ok, it can be done in a spreadsheet, I'll post it here, or if that's not possible I can send it to you.
Two questions: why do you reject cents as a tool for calculation; you must at least be using logarithms in your diagrams? And did you really find those other orders above 665 using the diagrammatic method?

Steve.

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

Hi Steve,

The way I did it was visually; the process is described in this image: http://i.imgur.com/dkySimH.png
That was my initial attempt; I later found out that 12 is not the division of the 1st order, 2 would be that:

Order # Generating steps

1 2 2 5 3 12 4 53 5 306 6 665 7 15.601 8 31.867 9 190.537 10 10.781.274 11 53.715.833
So my question here was how to determine the next steps, without having to draw 306 tiny logarithmic ticks on a number line. The algorithm should not include cents as a measuring tool; this is strictly ratios.

Your algorithm sound interesting, but I have no way of putting it into practice. I have no programming / technical skills, and I am just looking for something very simple, just like the picture I drew. Maybe something that could be calculated in a spreadsheet?

Bogdan
(baros_ilogic)

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi baros_ilogic,
As you may guess from the replies you've received, a lot is known about this process, and there is more than one number series that is related to this matter. Here's an interesting one:
2,3,5,7,12,17,29,41,53,94,147,200,253,306,359,665 ...
I see that Geddy is planning to say more about MOS; meanwhile here is my answer - I hope it will be helpful.

First, though, I am curious as to exactly how you found 53 as the next number after 12; you say that the leftover (Pythagorean comma) is about 1/53 of an octave, but actually it is nearer 1/51 of an octave, so I don't think you can have used it directly. I guess that you simply prolonged the chain of 5ths until you found a leftover that is "a lot smaller than" the small interval; I think that would work, but you would have a long way to go to get to 665, which I think is the next number. All the numbers in the list I gave above are MOS for the 3/2 interacting with the 2/1, but not all of them have a "small" leftover.

Here's an algorithm: given a MOS with N of interval A and M of interval B, find the next MOS by:
if A is the larger interval, then it's N of interval (A-B) and (N+M) of interval B;
if B is the larger interval, then it's M of interval (B-A) and (N+M) of interval A.
Hence define a new N, A, M and B. Repeat ad infinitum.
Note that (A-B) means the difference in cents or equivalently the quotient A/B if you're using fractions.
Start with the two-note MOS; i.e. 1 of interval 3/2 and 1 of interval 4/3.
You will want to do a further check of "evenness"; I suggest A/B (NB here I do mean a quotient of the cents values) between 3/4 and 4/3.

Steve M.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

I've uploaded a slightly improved version of the spreadsheet, as I realised that seven cells are inputs to the calculation, they are all highlighted in yellow now.

Steve.

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi Bogdan,
OK I understand; I agree that it's nice to see the powers of 2 and 3, but I can't easily perform the algorithm without cents (see below).
I have put a file called bogdan.xlsx in the Files section of this group; I hope you can read it. The "inputs" are at cells A3:D3, intermediate results below that in columns A:D, and the final results at cells R3:R50. Let me know if you want any further explanation.
Note that column N says if the First Interval is greater than the Second Interval; it is based on the cents values in Columns O and P.
Columns Q and S are alternative attempts to do the same test without using cents, but both fail after row 19. There may be a way to re-arrange the tests so that they work, but I have not spent the time to do it.

Steve.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Hi Steve,

Thanks for your interest in the topic. I don't know what to answer about cents... I'm trying to make an idea about the numbers, using simple and intuitive terms. For example, 3^53 * 2^-84 tells me much more than 3,615045866 cents. The place of the minus sign tells me that the comma appears on the left side of the graphic (meaning the comma is the first interval after 1/1) and is the difference between 53 twelfths and 84 octaves, or between 53 fifths and 31 octaves (because 84-53=31). And if there's small numbers we're after, the resulting decimal number 1,002090314 is as useful as its value in cents when spotting the next smaller comma. The next one after Mercator will be 3^-306 * 2^485; the minus sign is now with the 3, meaning that the comma is at the right side of the graphic (the comma is the last interval before 2/1).

Guess my understanding process has a preference for this kind of thinking. For example I couldn't understand MoS and Horograms until I got rid of cents and did all the calculations with ratios, like I explained here: http://groups.yahoo.com/neo/groups/tuning-math/conversations/messages/21381

The diagrams do use logarithms - they are actually ticks on a logarithmic number line. I only found the intervals up to 53 using the diagram method.

Bogdan

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi Bogdan,
Ok, it can be done in a spreadsheet, I'll post it here, or if that's not possible I can send it to you.
Two questions: why do you reject cents as a tool for calculation; you must at least be using logarithms in your diagrams? And did you really find those other orders above 665 using the diagrammatic method?

Steve.

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

Hi Steve,

The way I did it was visually; the process is described in this image: http://i.imgur.com/dkySimH.png
That was my initial attempt; I later found out that 12 is not the division of the 1st order, 2 would be that:

Order # Generating steps

1 2 2 5 3 12 4 53 5 306 6 665 7 15.601 8 31.867 9 190.537 10 10.781.274 11 53.715.833
So my question here was how to determine the next steps, without having to draw 306 tiny logarithmic ticks on a number line. The algorithm should not include cents as a measuring tool; this is strictly ratios.

Your algorithm sound interesting, but I have no way of putting it into practice. I have no programming / technical skills, and I am just looking for something very simple, just like the picture I drew. Maybe something that could be calculated in a spreadsheet?

Bogdan
(baros_ilogic)

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi baros_ilogic,
As you may guess from the replies you've received, a lot is known about this process, and there is more than one number series that is related to this matter. Here's an interesting one:
2,3,5,7,12,17,29,41,53,94,147,200,253,306,359,665 ...
I see that Geddy is planning to say more about MOS; meanwhile here is my answer - I hope it will be helpful.

First, though, I am curious as to exactly how you found 53 as the next number after 12; you say that the leftover (Pythagorean comma) is about 1/53 of an octave, but actually it is nearer 1/51 of an octave, so I don't think you can have used it directly. I guess that you simply prolonged the chain of 5ths until you found a leftover that is "a lot smaller than" the small interval; I think that would work, but you would have a long way to go to get to 665, which I think is the next number. All the numbers in the list I gave above are MOS for the 3/2 interacting with the 2/1, but not all of them have a "small" leftover.

Here's an algorithm: given a MOS with N of interval A and M of interval B, find the next MOS by:
if A is the larger interval, then it's N of interval (A-B) and (N+M) of interval B;
if B is the larger interval, then it's M of interval (B-A) and (N+M) of interval A.
Hence define a new N, A, M and B. Repeat ad infinitum.
Note that (A-B) means the difference in cents or equivalently the quotient A/B if you're using fractions.
Start with the two-note MOS; i.e. 1 of interval 3/2 and 1 of interval 4/3.
You will want to do a further check of "evenness"; I suggest A/B (NB here I do mean a quotient of the cents values) between 3/4 and 4/3.

Steve M.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

Hi Steve,

This looks great! Columns Q and S failed after row 19 because the numbers at numerator and denominator are too large for the spreadheet to handle. Nevertheless, the algorithm is excellent. I will need some time to figure out the meaning and use of columns A-D and all other yellow values, and will get back in case there are questions.

Bogdan

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi Bogdan,
OK I understand; I agree that it's nice to see the powers of 2 and 3, but I can't easily perform the algorithm without cents (see below).
I have put a file called bogdan.xlsx in the Files section of this group; I hope you can read it. The "inputs" are at cells A3:D3, intermediate results below that in columns A:D, and the final results at cells R3:R50. Let me know if you want any further explanation.
Note that column N says if the First Interval is greater than the Second Interval; it is based on the cents values in Columns O and P.
Columns Q and S are alternative attempts to do the same test without using cents, but both fail after row 19. There may be a way to re-arrange the tests so that they work, but I have not spent the time to do it.

Steve.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Hi Steve,

Thanks for your interest in the topic. I don't know what to answer about cents... I'm trying to make an idea about the numbers, using simple and intuitive terms. For example, 3^53 * 2^-84 tells me much more than 3,615045866 cents. The place of the minus sign tells me that the comma appears on the left side of the graphic (meaning the comma is the first interval after 1/1) and is the difference between 53 twelfths and 84 octaves, or between 53 fifths and 31 octaves (because 84-53=31). And if there's small numbers we're after, the resulting decimal number 1,002090314 is as useful as its value in cents when spotting the next smaller comma. The next one after Mercator will be 3^-306 * 2^485; the minus sign is now with the 3, meaning that the comma is at the right side of the graphic (the comma is the last interval before 2/1).

Guess my understanding process has a preference for this kind of thinking. For example I couldn't understand MoS and Horograms until I got rid of cents and did all the calculations with ratios, like I explained here: http://groups.yahoo.com/neo/groups/tuning-math/conversations/messages/21381

The diagrams do use logarithms - they are actually ticks on a logarithmic number line. I only found the intervals up to 53 using the diagram method.

Bogdan

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi Bogdan,
Ok, it can be done in a spreadsheet, I'll post it here, or if that's not possible I can send it to you.
Two questions: why do you reject cents as a tool for calculation; you must at least be using logarithms in your diagrams? And did you really find those other orders above 665 using the diagrammatic method?

Steve.

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

Hi Steve,

The way I did it was visually; the process is described in this image: http://i.imgur.com/dkySimH.png
That was my initial attempt; I later found out that 12 is not the division of the 1st order, 2 would be that:

Order # Generating steps

1 2 2 5 3 12 4 53 5 306 6 665 7 15.601 8 31.867 9 190.537 10 10.781.274 11 53.715.833
So my question here was how to determine the next steps, without having to draw 306 tiny logarithmic ticks on a number line. The algorithm should not include cents as a measuring tool; this is strictly ratios.

Your algorithm sound interesting, but I have no way of putting it into practice. I have no programming / technical skills, and I am just looking for something very simple, just like the picture I drew. Maybe something that could be calculated in a spreadsheet?

Bogdan
(baros_ilogic)

---In tuning-math@yahoogroups.com, <martinsj@...> wrote:

Hi baros_ilogic,
As you may guess from the replies you've received, a lot is known about this process, and there is more than one number series that is related to this matter. Here's an interesting one:
2,3,5,7,12,17,29,41,53,94,147,200,253,306,359,665 ...
I see that Geddy is planning to say more about MOS; meanwhile here is my answer - I hope it will be helpful.

First, though, I am curious as to exactly how you found 53 as the next number after 12; you say that the leftover (Pythagorean comma) is about 1/53 of an octave, but actually it is nearer 1/51 of an octave, so I don't think you can have used it directly. I guess that you simply prolonged the chain of 5ths until you found a leftover that is "a lot smaller than" the small interval; I think that would work, but you would have a long way to go to get to 665, which I think is the next number. All the numbers in the list I gave above are MOS for the 3/2 interacting with the 2/1, but not all of them have a "small" leftover.

Here's an algorithm: given a MOS with N of interval A and M of interval B, find the next MOS by:
if A is the larger interval, then it's N of interval (A-B) and (N+M) of interval B;
if B is the larger interval, then it's M of interval (B-A) and (N+M) of interval A.
Hence define a new N, A, M and B. Repeat ad infinitum.
Note that (A-B) means the difference in cents or equivalently the quotient A/B if you're using fractions.
Start with the two-note MOS; i.e. 1 of interval 3/2 and 1 of interval 4/3.
You will want to do a further check of "evenness"; I suggest A/B (NB here I do mean a quotient of the cents values) between 3/4 and 4/3.

Steve M.

---In tuning-math@yahoogroups.com, <baros_ilogic@...> wrote:

Confirming that this is the series I was looking for. This is the first time I hear about Pythagorean MOS.

How do you do this? I mean, in the innocent sense... How would you explain it to a 10-year old? Do you draw a horogram?

---In tuning-math@yahoogroups.com, <tuning-math@yahoogroups.com> wrote:

baros_ilogic wrote:

>
> Thank you Geddy; some of the things you describe exceed my knowledge... What I'm after is quite simple, no EDO's, no cents. The best way I can explain what it is in this picture. http://i.imgur.com/dkySimH.png

Sorry, seems I made a wrong assumption. Nonetheless, I think it still works
if we make a few slight changes, e.g. replacing "EDO" with "pythagorean MOS".

Can you confirm that the following sequence of pythagorean MOS scales is what
you're looking for?: 2 5 12 53 306 665 15601 31867 ...

pythagorean[2]: chroma = 9/8 (occurs as step in pythagorean[5])
pythagorean[5]: chroma = 256/243 (occurs as step in pythagorean[12])
pythagorean[12]: chroma = 531441/524288 (occurs as step in pythagorean[53])
pythagorean[53]: chroma = 3^53/2^84 (should occur as step in pythagorean[306])
pythagorean[306]: chroma = 2^485/3^306 (should occur as step in pythagorean[665])

pythagorean[665]: chroma = (something with 3^665) (should occur as step in pythagorean[15601])
pythagorean[15601]: chroma = (something with 3^15601) (should occur as step in pythagorean[31867])
...

The chroma of an MOS is the difference between the large and the small step.
With pythagorean[n] I mean the scale generated by stacking n fifth 3/2 modulo octave.
What I did was using the sequence of denominators of convergents for approximating the fifth
as a basis, but removed ones if still all subsequent convergents in the sequence are farey pairs
(so far I could only remove 41).
The farey pair relation is important to guarantee that the chroma of an MOS is a step in the
subsequent MOS (either the small or the large one), which was one of your requirements,
if I understood you correctly.

Removing as many MOS scales from the sequence as possible should make for more even
MOS scales, where the small and the large step are more similar in size.

Is this what you were looking for?

-Geddy

P.S.: I wrote earlier:
>
> For 53, you can use 94, 147, 200, 253, 306 or 359 (the two latter being convergents),
> or also 65, 118 etc.

Sorry, I made two mistakes here; I meant "denominator of convergent" instead of "convergent",
and it only applies to 306, but not to 359 (I probably looked in the wrong list).

that consist in:
((14337/14336)×(23247/23246)×(65601/65600)×(72225/72224)×(12627/12626)×
(18585/18584)×(62721/62720)×(76545/76544))^2×(676/675) = 3^53/2^84
as they do turn out during tempering an cycle of 53 quintes
by the following way:

+-0 : D. 1/1 unison
+1: A. 3/2
+2: E. 9/8
+3: B. 27/16
+4: F# 81/64
14336/14337 ~-0.1207570920216703609588013275859584129771296422405079549...cents
+5: C# 112/59 = (243/128)*[14336/14337]
+6: G# 84/59
+7: D# 63/59
+8: A# 189/118
23246/23247 ~-0.0744728905140009980255994927598659483594256730814373830...cents
+9: F/ (567/472)*[23246/23247] = 197/164 = (400/333)*[65601/65600]
65600/65601 ~-0.12075709202167036095880132758595841297712964224050795495...cents
+10: C/ 200/111
+11: G/ 50/37
+12: D/ 75/74 < 3^12/2^19
+13: A/ 225/148
72224/72225 ~-0.02397017700495556373690985638884653794161668354938...cents
+14: E/ (675/592)*[72224/72225] = 122/107 = (236/207)*[12627/12626]
12626/12627 ~-0.137111159067966002187946145554765558097725008047774785916...cents
+15: B/ 118/69
+16: F& 59/46
18584/18585 ~-0.137111159067966002187946145554765558097725008047774785916...cents
+17: C& (177/92)*[18584/18585] = 202/105
+18: G& 101/70 = (896/621)*[62721/62720]
62720/62721 ~-0.02760236364670928731935065205129291148229103726399086161...cents
+19: D& 224/207 = (303/280)*[62720/62721]
+20: A& 112/69
+21: F+ 28/23
+22: C+ 21/23
+23: G+ 63/46
+24: D+ 189/184 = (416/405)*[76544/76545]
76544/76545 ~-0.02261735394563135700054712388756123675681472539566786729...cents
+25: A+ 208/135 = (567/368)*[76545/76544]
+26: E+ 52/45 F-
675/676 ~-2.56289321278929683012354122939150709732015334656440032018...cents
-26: C- 45/26 B+
-25: G- 135/104
76544/76545
-24: D- 368/189
-23: A- 92/63
-22: E- 23/21
-21: B- 23/14
-20: GB 69/56
-19: DB 207/112
62720/62721
-18: AB 140/101
-17: EB 105/101
18584/18585
-16: BB 92/59
-15: F\ 69/59
12626/12627
-14: C\ 107/61
72224/72225
-13: G\ 296/225
-12: D\ 148/75
-11: A\ 37/25
-10: E\ 111/100
65600/65601
-9: B\ 328/197
23246/23247
-8: Gb 236/189
-7: Db 118/63
-6: Ab 59/42
-5: Eb 59/56
14336/14337
-4: Bb 128/81
-3: F. 32/27
-2: C. 16/9
-1: G. 4/3
+-0: D. 1/1 returned back to the unison

That circle yields in ascending order in extended Bosanquet-Helmholtz notation:

0: D. 1/1 unison
1: D/ 75/74
2: D+ 189/184 D//
3: EB 105/101 Eb\
4: Eb 59/56
5: D# 63/59
6: D& 224/207 D#/
7: E- 23/21 E\\
8: E\ 111/110
9: E. 9/8
10: E/ 122/107
11: E+ 52/45 F-
12: F\ 69/59
13: F. 32/27
14: F/ 197/164
15: F+ 28/23
16: GB 69/56
17: Gb 236/189
18: F# 81/64
19: F& 59/46
20: G- 135/104
21: G\ 296/225
22: G. 4/3
23: G/ 50/37
24: G+ 63/46
25: AB 140/101
26: Ab 59/42
27: G# 84/59
28: G& 101/70
29: A- 92/63
30: A\ 37/25
31: A. 3/2
32: A/ 225/148
33: A+ 208/135
34: BB 92/59
35: Bb 128/81
36: A# 189/118
37: A& 112/69
38: B- 23/14
39: B\ 328/197
40: B. 27/16
41: B/ 118/69
42: B+ 45/26 C-
43: C\ 107/61
44: C. 16/9
45: C/ 200/111
46: C+ 42/23
47: DB 207/112
48: Db 118/63
49: C# 112/59
50: C& 202/105
51: D- 368/189
52: D\ 148/75
53: D. 2/1

That sounds almost alike an bare Pythagorean chain of 52 consecutive 5ths.
The pure Pythagorean tuning consists in exactly 52 pure 5ths,
In the strict Pythagorean case,
most remote enharmonic 5th inside (E+=F-) and (B+=E-) amounts
the remaining
http://www.tonalsoft.com/enc/m/mercator-comma.aspx http://www.tonalsoft.com/enc/m/mercator-comma.aspx
http://xenharmonic.wikispaces.com/Mercator%27s+comma http://xenharmonic.wikispaces.com/Mercator%27s+comma
of ~-3.615...cents.
But here in the above presented case,
the Mercator-Comma got reduced downwards to
to the so called "Island-Comma" of:
http://xenharmonic.wikispaces.com/List+of+Superparticular+Intervals http://xenharmonic.wikispaces.com/List+of+Superparticular+Intervals
(see there in the list under 13-limit) quote:
" 676/675 http://xenharmonic.wikispaces.com/676_675 ~
2.5629....cents

island comma

"
The remaining difference of about ~1.052...cents got distributed
over the 16 other exiguos tiny epimoric subfactors.

bye
Andy